Why the NF and conversion loss is the same for passive fitle

Because the noise output is still ktB and the signal declines by the attenuation. The SNR out is worse than the SNR in by the attenuation.

the noise doesn't attenuate by the filter?

You'd better calculate the NF of a passive components.
That is NF=SNRin/SNRout=vnout^2/(vnin^2*A^2).
It's not a complex calculation.

The source noise is attenuated, but the circuit adds its own noise so it is back to kTB at the output.

KTB noise is everywhere. Your coffee cup has KTB noise. So, the output connector of the filter has its own noise.

I just don't understand why PNout=PNin.....

Added after 3 minutes:

I am sorry but i just know the KTB noise is the available noise power of a resistor.
Why the input noise power and the output noise power are both equal to KTB?They are independent to oter perameters and just rely to T&B?

I think u can look at it this way...the filter is usually at the beginning of the front-end..it has a certain attenuation, so it attenuates both the noise and signal, so ideally NF shud have been 1, but as the noise cannot do below kTB (physical phenomenon), noise at o/p of filter is still kTB...hence NF is equal to conversion loss

what if the filter is placed after the LNA or mixer?

Let me attempt to clarify the issue. If the passive netwrok is "ideally" noiseless, then the NF would be equal to the loss. This has nothing to do with the noise added by the filter. What the passive filter does is simply shape the noise spectral density. However, the total noise power will remain the same at the output. If the passive network is noisy, in the case of a passive mixer for example, the NF can be higher than the loss.

The NF is the input s/n to he output s/n in a certain bandwidth, if the passive filter just shape the noise spectral density, the output noise power in the bandwidth will still be different from the input. Moreover, I don't think what the passive filter do is just shape the noise spectral density, it does attenuate the noise power.

"What the passive filter does is simply shape the noise spectral density. However, the total noise power will remain the same at the output."

If it shapes the noise spectral density (by filtering),
how come the noise at the output is still the same?

is it because you cant get below KTB ?

The Noise at the I/P is the same as the O/P for Passive circuits!
But if the Signal Power is attenuated (i/p to o/p) by the I.L. Then :

F= (S/N)in / (S/N)out

You see that the signal attenuation is the only contribution to the Noise Definition.

If the noise is much higher than KTB, why cant a passive filter attenuate the noise, just like it does to the signal?

If NF is the same than attenuation in an ideal filter, doesn't it comes from impedance matching at input/output? I mean isn't that only true when the same R is seen at input and output (from kTBR)?

Cheers

why the noise at the I/P is the same as the o/p for passive circuits?

I believe that if you have a passive netwrok, with a source impedance Rs connecting the input terminals, then you can represent the netwrok with a Thevenin noise source Vn and an output resistance Rout, where:

Vn^2=4kTRoutB

Note that Rout will be dependent on Rs